Solve: 10^logx + log4 ln(2x^2 + 4x) = 5 + ln2x

To solve 

(i)  10^logx
+log4 

(ii)
ln(2x^2+4x)=5+ln2x.

(i) 10^logx +log4 is not an expression
(not an equation). It could be simplified only.

First term:
Let 10^logx = y. So by definition logx = log y. Or y = x. Therefore 10^logx =
x.

So 10 ^logx +log4 = x+log4 .If this expression is equal
to a number n,

x+log 4 = n.
So

x = n- log4.

ln(2x^2+4x) =
5+ln2x.

ln(2x^2+4x) -ln2x =
5

ln{(2x^2+4x)/2x} = 5,as lna-lnb =
ln(a/b)

ln(x+2) = 5 =
lne^5.

x+2 = e^5, as

x =
(e^5)-2